3/21/2015
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Practice Exam # 2 (2.7­3.9) (6968235)
Due:
Thu Mar 26 2015 11:59 PM PDT
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Question
1.
Question Details
SCalcET7 2.8.027. [1735792]
­
Find the derivative of the function using the definition of derivative.
g(x) = 1 − x
g'(x) = State the domain of the function. (Enter your answer using interval notation.)
State the domain of its derivative. (Enter your answer using interval notation.)
Solution or Explanation
g'(x) = lim h → 0 g(x + h) − g(x)
h
1 − (x + h) − = lim h → 0 1 − (x + h) + 1 − x
1 − (x + h) + 1 − x
[1 − (x + h)] − (1 − x)
= lim h → 0 h
1 − (x + h) + 1 − x
−h
= lim h → 0 h
1 − (x + h) + 1 − x
−1
= lim 1 − (x + h) + h → 0 = 1 − x
h
1 − x
−1
2
1 − x
Domain of g = (−∞, 1], domain of g' = (−∞, 1).
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2.
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SCalcET7 2.8.037. [1637474]
­
SCalcET7 3.1.069. [1783058]
­
The graph of f is given. State the numbers at which f is not differentiable.
x=
­4 (smaller value)
x=
0 (larger value)
Solution or Explanation
Click to View Solution
3.
Question Details
Consider the following function.
f(x) = |x2 − 9|
(a) Find a formula for f '.
if |x| > 3
if |x| < 3
f '(x) = For what values of x is the function not differentiable? (Enter your answers as a comma­separated list.)
x = (b) Sketch the graph of f.
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Sketch the graph of f '
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Solution or Explanation
(a) Note that x2 − 9 < 0 for x2 < 9 ⇔ |x| < 3 ⇔ −3 < x < 3. So x2 − 9 if x ≤ −3
f(x) = −x2 + 9 if −3 < x < 3 x2 − 9 if x ≥ 3
2x if x < −3
2x if |x| > 3
f '(x) = −2x if −3 < x < 3 = −2x if |x| < 3
2x if x > 3
To show that f '(3) does not exist we investigate lim f(3 + h) − f(3)
h
h → 0 [−(3 + h)2 + 9] − 0
by computing the left­ and right­hand derivatives.
f(3 + h) − f(3)
lim and
= lim = h → 0− (−6 − h) = −6
h
h
h → 0− 2
2
f(3 + h) − f(3)
lim [(3 + h) − 9] − 0 = lim 6h + h = lim (6 + h) = 6.
f ' +(3) = lim = h
+
+
h
h
h → 0+ h → 0 h → 0 h → 0+ f(3 + h) − f(3)
Since the left and right limits are different, lim does not exist, that is, f '(3) does not exist. Similarly, h
h → 0 f ' −(3) = lim h → 0− f '(−3) does not exist. Therefore, f is not differentiable at 3 or at −3.
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4.
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Question Details
SCalcET7 3.2.502.XP. [1836132]
­
SCalcET7 3.2.020. [1835633]
­
SCalcET7 3.3.015. [1835932]
­
Find the derivative of the function below in two ways.
F(x) = x − 3x
x
x
(a) by using the Quotient Rule
F'(x) = (b) by simplifying first
F'(x) = (c) Which method appears to be simpler for this problem?
the Quotient Rule
simplifying first Solution or Explanation
Click to View Solution
5.
Question Details
Differentiate. (Assume c is a constant.)
z = u3/2(u + ceu)
z' = Solution or Explanation
Click to View Solution
6.
Question Details
Differentiate.
f(x) = 2xex csc x
f '(x) = Solution or Explanation
Click to View Solution
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7.
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SCalcET7 3.3.022. [1836395]
­
SCalcET7 3.4.023. [1799833]
­
Find an equation of the tangent line to the curve at the given point.
y = 3ex cos x, (0, 3)
y = Solution or Explanation
Click to View Solution
8.
Question Details
Find the derivative of the function.
y = 5 + 8e9x y' = Solution or Explanation
y = 5 + 8e9x y' = 1
d
(5 + 8e9x )−1/2
(5 + 8e9x ) = 2
2
dx
1
(8e9x · 9) = 5 + 8e9x
36e9x
5 + 8e9x
9.
Question Details
SCalcET7 3.4.031. [1835882]
­
SCalcET7 3.4.033. [1835657]
­
Find the derivative of the function.
y = sin(tan 6x)
y' = Solution or Explanation
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10.
Question Details
Find the derivative of the function.
y = 7sin πx
y' = Solution or Explanation
Click to View Solution
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11.
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SCalcET7 3.4.041. [1836073]
­
SCalcET7 3.4.044. [1836022]
­
SCalcET7 3.5.008.MI. [1645360]
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SCalcET7 3.5.019. [1836234]
­
Find the derivative of the function.
2
f(t) = cos2(ecos t)
f '(t) = Solution or Explanation
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12.
Question Details
Find the derivative of the function.
y' = Solution or Explanation
Click to View Solution
13.
Question Details
Find dy/dx by implicit differentiation.
6x3 + x2y − xy3 = 3
y' = Solution or Explanation
Click to View Solution
14.
Question Details
Find dy/dx by implicit differentiation.
ey cos x = 9 + sin(xy)
y' = Solution or Explanation
Click to View Solution
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15.
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Question Details
SCalcET7 3.6.004. [1836239]
­
SCalcET7 3.6.007. [1799716]
­
SCalcET7 3.6.011. [1835580]
­
Differentiate the function.
f(x) = ln(36 sin2x)
f '(x) = Solution or Explanation
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16.
Question Details
Differentiate the function.
f(x) = log10(x9 + 7)
f '(x) = Solution or Explanation
f(x) = log10(x9 + 7) f '(x) = 1
d
9x8
(x9 + 7) = (x9 + 7)ln 10 dx
(x9 + 7)ln 10
17.
Question Details
Differentiate the function.
g(x) = ln x
x2 − 5
g'(x) = Solution or Explanation
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18.
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Question Details
SCalcET7 3.6.041. [1799740]
­
Use logarithmic differentiation to find the derivative of the function.
y = x − 4
x8 + 5
y' = Solution or Explanation
y = x − 4
x8 + 5
y' = y
ln y = ln
1
4x7
− 8
2(x − 4)
x + 5
x − 4
x8 + 5
1/2
ln y = y' = x − 4
x8 + 5
1
1
ln(x − 4) − ln(x8 + 5) 2
2
1
1
1
1
1
y' =
− 8
· 8x7 y
2 x − 4
2 x + 5
1
4x7
− 8
2x − 8
x + 5
19.
Question Details
SCalcET7 3.6.044.MI. [1836262]
­
Use logarithmic differentiation to find the derivative of the function.
y = x4 cos x
y' = Solution or Explanation
Click to View Solution
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20.
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SCalcET7 3.7.007. [1815468]
­
The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of
21.5 m/s is h = 4 + 21.5t − 4.9t2 after t seconds. (Round your answers to two decimal places.)
(a) Find the velocity after 2 s and after 4 s.
v(2) = 1.9 m/s
v(4) = ­17.7 m/s
(b) When does the projectile reach its maximum height?
2.19 s
(c) What is the maximum height?
27.58 m
(d) When does it hit the ground?
4.57 s
(e) With what velocity does it hit the ground?
­23.25 m/s
Solution or Explanation
(a) h(t) = 4 + 21.5t − 4.9t2 v(t) = h'(t) = 21.5 − 9.8t. the velocity after 2 s is v(2) = 21.5 − 9.8(2) = 1.9 m/s
and after 4 s is v(4) = 21.5 − 9.8(4) = −17.7 m/s.
(b) The projectile reaches its maximum height when the velocity is zero. v(t) = 0 ⇔ 21.5 − 9.8t = 0 ⇔ 21.5
t = = 2.19 s.
9.8
(c) The maximum height occurs when t = 2.19. h(2.19) = 4 + 21.5(2.19) − 4.9(2.19)2 = 27.58 m.
(d) The projectile hits the ground when h = 0 ⇔ 4 + 21.5t − 4.9t2 = 0 ⇔ t = −21.5 ± 21.52 − 4(−4.9)(4) 2(−4.9)
t = tf ≈ 4.57 s [since t ≥ 0].
(e) The projectile hits the ground when t = tf. Its velocity is v(tf) = 21.5 − 9.8tf ≈ −23.25 m/s [downward].
21.
Question Details
SCalcET7 3.9.005.MI. [1646152]
­
A cylindrical tank with radius 7 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water
increasing? m/min Solution or Explanation
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22.
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SCalcET7 3.9.014. [2727122]
­
At noon, ship A is 170 km west of ship B. Ship A is sailing east at 40 km/h and ship B is sailing north at 15 km/h. How fast
is the distance between the ships changing at 4:00 PM? km/h
Solution or Explanation
Given: at noon, ship A is 170 km west of ship B; ship A is sailing east at 40 km/hr and ship B is sailing north at 15 km/h.
If we let t be time (in hours), x be the distance traveled by ship A (in km), and y be the distance traveled by ship B (in
km), then we are given that dx/dt = 40 km/h and dy/dt = 15 km/h. Unknown: the rate at which the distance between the ships is changing at 4:00 PM. If we let x be the distance between
the ships, then we want to find dz/dt when t = 4h. z2 = (170 − x)2 + y2 2z dz
dx
dy
= 2(170 − x) − + 2y dt
dt
dt
At 4:00 PM, x = 4(40) = 160 and y = 4(15) = 60 So z = (170 − 160)2 + 602 = 10
37 . (−10)(40) + 60(15)
50
dz
1
dx
dy
= (x − 170)
+ y
= = ≈ 8.2199 km/h.
10 37
dt
z
dt
dt
37
23.
Question Details
SCalcET7 3.9.017. [1646109]
­
A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point
500 ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking? (Round your answer
to two decimal places.)
4.98 ft/s
Solution or Explanation
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24.
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SCalcET7 3.9.028. [1647667]
­
A kite 100 ft above the ground moves horizontally at a speed of 4 ft/s. At what rate is the angle (in radians) between the
string and the horizontal decreasing when 200 ft of string have been let out?
1/100 rad/s
Solution or Explanation
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Name (AID): Practice Exam # 2 (2.7­3.9) (6968235)
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